[geocentrism] Re: Tides

  • From: Regner Trampedach <art@xxxxxxxxxxxxxx>
  • To: geocentrism@xxxxxxxxxxxxx
  • Date: Tue, 13 May 2008 00:25:45 +1000

Neville Jones wrote:
Regner,

Your analysis of the cause of the tides is invalid, because gravitational attraction between parts of extended objects being exactly equivalent to all mass of each object being concentrated into mathematical points is valid for these objects being rigid, not flexible. You cannot assume the World to be rigid on the one hand and then subtract the theoretical force on a rigid World from both sides of a flexible World in order to 'explain' the tides.
There is no such assumption in my analysis.
Close to a body, you have to take into account the mass distribution in that bode, in order to
get an accurate value of the gravitational force from that object.
To calculate an accurate value of the gravitational acceleration by the Earth on you, we would
need to know the detailed mass-distribution inside the Earth - considering the Earth a point-object
in this case, would be a bad approximation.
  The gravitational force on you by the Moon, however, is accurately given by approximating the
Moon by a point-mass.
  Likewise, the difference in gravitational force by the Moon on opposite sides of the Earth, are
accurately given by approximating the Moon by a point-mass.
  If the Earth was completely rigid, the tidal forces from the Moon would still be the same - they
have nothing to do with the composition of the Earth - only the distance from the Moon and
the diameter of the Earth enters here.
  The response of the Earth to this tidal force, obviously depends on the composition of the Earth,
but I know too little about that to be able to calculate anything beyond orders-of-magnitude
estimates. Notice, however, that my post didn't say anything about the amplitude or phase of the
tides on Earth - I only gave it's cause, and the cause of tides is what was misunderstood by other
posters here.
  The acceleration of the centre-of-mass of the Earth behaves exactly as if all the Earth's mass was
in that point. That can be fairly easily proven mathematically (and is what we observe). The same
obviously goes for the Moon. So the orbits of the Earth and the Moon around their common bary-
centre, can be accurately described in the point-mass approximation.
  If you subtract that motion and multiply by the mass, you can get the tidal forces on opposite
sides of the Earth. There is nothing inconsistent here.

Where would these forces on the oceans be acting? At the centre of mass of the ocean, or on individual water molecules, or at the centre of mass of the World?
Gravity acts between each and every particle with mass.

The pull would vary as the inverse-square of the distance, which would result in very serious tidal waves arising as the Moon-side ocean was pulled more than the other was 'left behind'.
Well, this is what we experience as tides. If you want to call that a "...very serious tidal wave...",
be my guest. Remember that Earth's own gravity is still pulling every part of the Earth towards
it's own centre-of-mass (Earth is self-gravitating).
   The gravitational acceleration by the Moon at the surface of the Earth, is about a millionth of that due
to the Earth in the same spot. The tidal acceleration is another factor of 30 smaller. It is not that hard to
actually put some numbers in, and check ones assumptions.
   Incidentally, the centrifugal force produced by Earth's daily rotation is about 30,000 times larger than
the tidal force from the Moon, and produces the (stationary) 21.3km equatorial bulge. Assuming the
displacement to be proportional to force (as with an ideal spring/elastic band), the Lunar tidal force
would result in a 7mm difference between high- and low-tide in the Earth's crust. The observed effect
is a few centimeters, and obviously depends on the composition of the local crust - orders-of-magnitude
estimates can be pretty powerful (this one wasn't far off)...

Rather than being   [I moved these a bit to align with my equations below/Regner]
                 <-                            |                               ->    as depicted, the force would be proportional to

        x*x^(-1)          (x*x+2xR+R*R)^(-1)   (x*x+4xR+4R*R)^(-1)
It is the same thing, Neville. You give the total gravitational force by the Moon on
various parts of the Earth in those expressions. Now let me rephrase in terms of the
distance, d, between the centre-of-mass of the Earth and the Moon (d=x+R) to make
it symmetric. Then we get:
     (d-R)^(-2)                             d^(-2)               (d+R)^(-2)
which is the same as
    (d*d-2dR+R*R)^(-1)            d^(-2)        (d*d+2dR+R*R)^(-1)
Just substitute   d=x+R  and you'll see it is the same. Now since, R << d, we can
Taylor expand in (R/d). First we prepare our _expression_ by taking the leading
term, d^(-2), outside, to get
    [d*d(1-2R/d+(R/d)^2]^(-1)    d^(-2)      [d*d(1+2R/d+(R/d)^2)]^(-1)
Since (R/d) is very small, we can to the first approximation, neglect (R/d)^2 so that
    [d*d(1-2R/d)]^(-1)                 d^(-2)      [d*d(1+2R/d]^(-1)
Again, since (R/d) is very small, we can approximate 1/(1+2R/d)  by  1-2R/d - that
is the Taylor expansion, and it gives us
    1/d^2 + 2R/d^3                      1/d^2       1/d^2 - 2R/d^3
If we now subtract the motion of the centre-of-mass of the Earth, 1/d^2, then we are
left with the tidal forces:
               + 2R/d^3                        0                     - 2R/d^3
-2/d^3 is the derivative of  1/d^2  with respect to d, and then we multiply by the radius
of the Earth, R, to get the tidal force - exactly as I wrote in my previous post.
  In our calculations, we can just as well use the exact expressions, before the Taylor
expansion and before throwing away the 2nd order term in (R/d). The difference is
very small, however, and the form I wrote, conveys a lot more meaning to a physicist.

where x is the distance from the centre of mass of the Moon to the closest tip of the ocean nearest to the Moon and R is the radius of the World, and x >> R in the heliocentric model.
I'm a bit confused. Why does x and R change if the Universe is Geocentric?!?

Ocean swell would also vary with the Moon's orbit.

As they do.
The analysis is thus far too simplistic.
Yes - and I never did say that I provided a thorough analysis of the tidal interactions
between the Earth and the other bodies in the Solar system. I gave a simple analysis
aimed at being understood by laymen, and which explains the major forces that
govern tides on Earth.
As with other aspects of the so-called 'solar system', such forces would lead very quickly to disorder and destruction.

Nope. Why would they? See the actual numbers in green, above.

This reply got much longer than I had intended - sorry.

      Regner
Neville.


-----Original Message-----
From: art@xxxxxxxxxxxxxx
Sent: Wed, 07 May 2008 15:18:40 +1000

There has been a couple of references to tides, lately, and much confusion
on the subject. I'll do what I can to clear it up a bit.
Please take the time to read the whole post before commenting.

The force of gravity moves objects
The Moon and the Earth orbit each other due to their gravitational
interaction.
It can be shown that you can simplify this problem:
   Every single bit of the Moon interacting gravitationally with every single
   bit of the Earth.
is exactly equivalent to
   All the mass of the Moon being located at the centre-of-mass of the Moon,
   and all the mass of the Earth being located at the centre-of-mass of the
   Earth.
From such an analysis we find that the two objects will orbit their common
centre-of-mass in elliptical orbits.

Tidal forces deform objects
But the Earth and the Moon are extended objects so what effect does that
have?  Well, the Moon is still close enough to the Earth, that the gravity
from the Moon will change over the diameter of the Earth:

    Moon                                Near     Centre     Far
                                        side    of Earth    side
      O                               <-------- <-------  <------
 
Okay, if we now subtract the part that actually moves the Earth, that is
we subtract   <------- which is the same as adding ------->  to all three
vectors (arrows) above, and we get the tidal forces:
    Moon                                Near     Centre     Far
                                        side    of Earth    side
      O                                  <-         |        ->
So looking at the Moon-Earth system from the outside we see that the
Moon-side of Earth experience a larger gravitational pull from the Moon
than the centre of the Earth, and the opposite side of Earth experience
experience even less gravitational pull from the Moon. Seeing it from the
Earth point of view, we have a Moon-ward force on the Moon-side of Earth
and an opposite force on the opposite side of Earth. This is what causes
the tidal bulge, and this is why there are (approximately) two high- and two
low-tides a day. The oceans, being a liquid, obviously responds much stronger
than the bedrock, but the Earth's crust is flexing too, as has been measured.
The Earth rotates under this tidal bulge, which makes it a dynamical phenomenon.
  The gravitational force is proportional to  1/r²  where r is the distance
between the Earth and the Moon. The tidal force is the difference in
gravitational force over a certain distance, which means it can be expressed
as the r-derivative of the gravitational force. The tidal force is therefore
proportional to  1/r³. That means the tidal force falls off more rapidly with
distance than the gravitational force. This is the reason that the gravitational
force of the Sun dominates the orbit of the Earth, but the Moon's tidal forces
on Earth are larger than the Sun's.
  The Sun also contributes a significant tide (about half that of the Moon)
and when the two are aligned at new Moon or full Moon, we get more
powerful tides called spring tides. At either half Moon we get neap tides.
The ellipticity (eccentricity) of the Moon-Earth and the Sun-Earth orbits
also causes variations in the strength of tides on Earth.
  The Moon experience (about 22 times) stronger tides from the Earth
because of the Earth/Moon mass ratio of 81 and the size ratio of 3.7.
The Lunar tide from the Sun is 3.7 times weaker than here on Earth, due
to the 3.7 times smaller diameter of the Moon.

The tides would of course be exactly the same in a geocentric Universe.

Allen's concerns in the "Acceleration thread"
To account for tidal influences, the centre-of-mass of the test-mass of
the accelerometer, should coincide with the centre-of-mass of the free-
falling laboratory. Tides would have NO effect on such a set-up.
But then the tidal effects on such a lab are very small in the first place:
Less than a billionth per meter of the gravitational force keeping the lab
AND the test-mass in orbit.

That's all for now (sorry for such a long post), and I hope it clarified a
few things.

          - Regner
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